Question : If $3 \tan \theta=2 \sqrt{3} \sin \theta, 0^{\circ}<\theta<90^{\circ}$, then the value of $\frac{\operatorname{cosec}^2 2 \theta+\cot ^2 2 \theta}{\sin ^2 \theta+\tan ^2 2 \theta}$ is:
Option 1: $\frac{4}{13}$
Option 2: $\frac{20}{39}$
Option 3: $\frac{4}{3}$
Option 4: $\frac{20}{27}$
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Correct Answer: $\frac{20}{39}$
Solution : $3 \tan \theta=2 \sqrt{3} \sin \theta$ ⇒ $3 × \frac{\sin \theta}{\cos \theta} = 2 \sqrt{3} \sin \theta$ ⇒ $\frac{3}{\cos \theta} = 2 \sqrt{3}$ ⇒ $\cos \theta = \frac{3}{2 \sqrt{3}} = \frac{\sqrt3}{2}$ ⇒ $\theta = 30^{\circ}$ Now, $\frac{\operatorname{cosec}^2 2 \theta+\cot ^2 2 \theta}{\sin ^2 \theta+\tan ^2 2 \theta}$ $=\frac{\operatorname{cosec}^2 (2 ×30^{\circ})+\cot ^2 (2 ×30^{\circ})}{\sin ^2 30^{\circ}+\tan ^2 (2× 30^{\circ})}$ $=\frac{\operatorname{cosec}^2 60^{\circ}+\cot ^2 60^{\circ}}{\sin ^2 30^{\circ}+\tan ^2 60^{\circ}}$ $=\frac{(\frac{2}{\sqrt3})^2+(\frac{1}{\sqrt3})^2}{(\frac{1}{2})^2+(\sqrt3)^2}$ $=\frac{(\frac{4}{3})+(\frac{1}{3})}{(\frac{1}{4})+3}$ $=\frac{(\frac{5}{3})}{(\frac{1+12}{4})}$ $= \frac{5×4}{3×13}$ $=\frac{20}{39}$ Hence, the correct answer is $\frac{20}{39}$.
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Question : $\frac{1+\cos \theta-\sin ^2 \theta}{\sin \theta(1+\cos \theta)} \times \frac{\sqrt{\sec ^2 \theta+\operatorname{cosec}^2 \theta}}{\tan \theta+\cot \theta}, 0^{\circ}<\theta<90^{\circ}$, is equal to:
Question : $\left(\frac{\tan ^3 \theta}{\sec ^2 \theta}+\frac{\cot ^3 \theta}{\operatorname{cosec}^2 \theta}+2 \sin \theta \cos \theta\right) \div\left(1+\operatorname{cosec}^2 \theta+\tan ^2 \theta\right), 0^{\circ}<\theta<90^{\circ}$, is equal to:
Question : The expression $\frac{\cos ^4 \theta-\sin ^4 \theta+2 \sin ^2 \theta+3}{(\operatorname{cosec} \theta+\cot \theta+1)(\operatorname{cosec} \theta-\cot \theta+1)-2}, 0^{\circ}<\theta<90^{\circ}$, is equal to:
Question : If $\cos \left(2 \theta+54^{\circ}\right)=\sin \theta, 0^{\circ}<\left(2 \theta+54^{\circ}\right)<90^{\circ}$, then what is the value of $\frac{1}{\tan 5 \theta+\operatorname{cosec} \frac{5 \theta}{2}}$?
Question : The expression $(\tan \theta+\cot \theta)(\sec \theta+\tan \theta)(1-\sin \theta), 0^{\circ}<\theta<90^{\circ}$, is equal to:
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